Problem: You have found the following ages (in years) of all 4 bears at your local zoo: $ 10,\enspace 16,\enspace 28,\enspace 12$ What is the average age of the bears at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 4 bears at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{10 + 16 + 28 + 12}{{4}} = {16.5\text{ years old}} $ Find the squared deviations from the mean for each bear. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $10$ years $-6.5$ years $42.25$ years $^2$ $16$ years $-0.5$ years $0.25$ years $^2$ $28$ years $11.5$ years $132.25$ years $^2$ $12$ years $-4.5$ years $20.25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{42.25} + {0.25} + {132.25} + {20.25}} {{4}} $ $ {\sigma^2} = \dfrac{{195}}{{4}} = {48.75\text{ years}^2} $ The average bear at the zoo is 16.5 years old. The population variance is 48.75 years $^2$.